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Re:当best_t[i]!=tt[i]时就要break,因为大小显然只与第一个不同的数有关.

Posted by lbp at 2005-09-17 16:09:13 on Problem 1042
In Reply To:当best_t[i]!=tt[i]时就要break,因为大小显然只与第一个不同的数有关. Posted by:number at 2005-09-16 14:27:41

实际上，不管best_t[i] <tt[i] 还是best_t[i] <=tt[i]；紧接的break；必然要被执行；所以下面的i!=n+1必然为真，，，也就是best_t[i] <tt[i]与best_t[i] <=tt[i]在这 是等效的.......不会出现best_t[i] >=tt[i]对所有i成立的情况。。

Followed by:

没意思，有时间多看看算法 (164) semonteer 2005-09-17 16:21:35
<=,<,都能过,会出现best_[i]>=tt[i]对所有i成立的情况. (164) number 2005-09-17 16:25:37
- 出现best_[i]>=tt[i]对所有i成立的原因不在题目数据,而在算法. (58) number 2005-09-17 16:28:36

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